Page 2 of 4 FirstFirst 1234 LastLast
Results 16 to 30 of 46
Like Tree6Thanks

Math Help - Uniform continuity

  1. #16
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    Quote Originally Posted by Plato View Post
    Both of your proofs do not work.

    Here is the relevant theorem that you need.
    If is continuous on and then is uniformly continuous on
    Is this a theorem you have made up? If so, please prove it, or give a source for the theorem and its proof.

    The only theorem I have seen is f is uniformly continuous on [a,b] if f is continuous on [a,b].

    We are all aware of the epsilon/delata requirement.

    The OP can be proved for a finite closed interval.

    None of the proofs so far are correct. Do you have a proof?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    Is this a theorem you have made up? If so, please prove it, or give a source for the theorem and its proof.
    The only theorem I have seen is f is uniformly continuous on [a,b] if f is continuous on [a,b].
    None of the proofs so far are correct. Do you have a proof?
    The statement and proof of Theorem 4.15 appears on page 115 of Introduction to Mathematical Analysis by William R. Parzynski and Philip Zipse (McGraw-Hill).
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    Buck, Rosenlicht, Rudin, Spivak and Taylor give the theorem for a compact space (finite, closed interval). Referring to an obscure book which most people never heard of or don't have is quite convenient.

    Personally, I don't believe it, things get tricky when limits become infinite. I would be more inclined to believe it if someone could prove OP is uniformly continuous by basic definition. Perhaps you could?



    However, if the theorem is indeed true, then you have answered OP.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,937
    Thanks
    785

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    Is this a theorem you have made up? If so, please prove it, or give a source for the theorem and its proof.

    The only theorem I have seen is f is uniformly continuous on [a,b] if f is continuous on [a,b].

    We are all aware of the epsilon/delata requirement.

    The OP can be proved for a finite closed interval.

    None of the proofs so far are correct. Do you have a proof?
    Let $f$ be a function that is continuous on $[a,\infty)$ and $\lim_{x \to \infty} f(x) = k$.

    Given $\varepsilon>0$, choose $b \in [a,\infty)$ such that for all $x \ge b$, $\left| f(x) - k \right| < \dfrac{\varepsilon}{3}$ (we know $b$ exists since the limit as $x \to \infty$ of $f(x)$ exists).

    Since $f$ is continuous on the compact set $[a,b] \subseteq [a,\infty)$, by the Heine-Cantor theorem, $f$ is uniformly continuous on $[a,b]$. Choose $\delta>0$ such that given any $x,y \in [a,b]$ with $|y-x|< \delta$, $|f(y) - f(x)| < \dfrac{\varepsilon}{3}$.

    Let $x,y \in [a,\infty)$ with $|y-x| < \delta$. We have the following possibilities:

    1. $x\le b$ and $y\le b$
    2. $x\le b$ and $y> b$ or $x>b$ and $y\le b$
    3. $x>b$ and $y>b$

    Case 1:
    Since we already know that $f$ is uniformly continuous on $[a,b]$, we know $|f(y) - f(x)| < \dfrac{\varepsilon}{3} < \varepsilon$

    Case 2:
    Showing the case when $y>x$ (the case when $x>y$ is similar).

    By the triangle inequality, we know

    $|f(y) - f(x)| \le |f(y) - k| + |k - f(b)| + |f(b) - f(x)|$

    We chose $b$ such that for all $x\ge b$, $|f(x) - k| < \dfrac{\varepsilon}{3}$. Also, we know that since $x \le b < y$ (by assumption), then $|b-x| < |y-x| < \delta$, so we have

    $|f(y) - f(x)| \le |f(y) - k| + |k - f(b)| + |f(b) - f(x)| < \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} = \varepsilon$

    Case 3:

    $|f(y) - f(x)| \le |f(y) - k| + |k - f(x)| < \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} < \varepsilon$

    Q.E.D.
    Last edited by SlipEternal; February 19th 2014 at 08:04 AM.
    Thanks from emakarov
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    For uniform continuity, the proof can't depend on b.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,937
    Thanks
    785

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    For uniform continuity, the proof can't depend on b.
    It doesn't. It depends on $\varepsilon$.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    OK, given ε you can pick b. After that, I will check that the same δ applies to all cases. Intuitiveley, you shouldn't have to break it down by cases depending on where x and y are.

    Thanks for giving an intelligible proof which can be followed. It's interesting. It appears that it also applies to a finite interval which makes me wonder.

    I note derivative is not continuous on the interval but that is only sufficient, but not necessary (if your continuity proof is correct).

    Why doesn't someone nail it by answering OP using basic definition of uniform convergence?

    Edit: Whoops. Given ε, you can pick b, which depends on x. Maybe that's why the method isn't used in a standard proof for[a,b]
    Last edited by Hartlw; February 19th 2014 at 08:38 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    Think I'll reinforce my last edit:

    Given ε, δ depends on b, which depends on x. δ has to be independent of x.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,937
    Thanks
    785

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    Think I'll reinforce my last edit:

    Given ε, δ depends on b, which depends on x. δ has to be independent of x.
    I don't see the dependence on x. I suppose I will have to take your word for it.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    You don't have to take my word for it.

    Given ε>0 , choose b∈[a,∞) such that for all x≥b , |f(x)−k|<ε 3
    Follow Math Help Forum on Facebook and Google+

  11. #26
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,937
    Thanks
    785

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    You don't have to take my word for it.

    Given ε>0 , choose b∈[a,∞) such that for all x≥b , |f(x)−k|<ε 3
    So, $b$ depends on $\varepsilon$. I still don't see any dependence upon $x$. But again, if you say there is, I will take your word for it.
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    You have to choose b st if x>b, ...

    Please spare me the gratuitous insults. If you don't see it, I doubt very much you will take my word for it.
    Follow Math Help Forum on Facebook and Google+

  13. #28
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    Why doesn't someone nail it by answering OP using basic definition of uniform convergence?
    O.K. I will do that for you, even though I doubt you will follow it.

    Start every continuity proof with: SUPPOSE $\varepsilon > 0$.

    Note that $\displaystyle{\lim _{x \to \infty }}\left( {f(x) = {{\sin ({x^3})} \over x}} \right) = 0$.

    That means that $\left( {\exists N > a} \right)\left[ {x \geqslant N \Rightarrow \;\left| {f(x)} \right| < \dfrac{\varepsilon }{3}} \right]$

    Now $f$ is uniformly continuous on $[a,N]$ so $\left( {\exists {\delta _1} > 0} \right)\left[ {\left| {x - y} \right| < {\delta _1} \Rightarrow \;\left| {f(x) - f(y)} \right| < \frac{\varepsilon }{3}} \right]$.

    Also by continuity $\left( {\exists {\delta _2} > 0} \right)\left[ {\left| {N - y} \right| < {\delta _2} \Rightarrow \;\left| {f(N) - f(y)} \right| < \frac{\varepsilon }{3}} \right]$.

    Let $\delta = \min \left\{ {{\delta _1},{\delta _2}} \right\}$ so suppose that $\left| {p - q} \right| < \delta $.

    Three cases: $ \begin{array}{l} 1.~\{p,q\}\subset [a,N]\\2.~p<N< q\\3.~ \{p,q\}\subset [N,\infty)\end{array} $

    In each case $\left| {f(p) - f(q)} \right| < \varepsilon $.

    BTW: Philip Zipse is hardly obscure.
    Last edited by Plato; February 19th 2014 at 09:31 AM.
    Thanks from emakarov
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Uniform continuity

    The function f(x) = x2 is contiuous but not uniformly continuous on (0,infinity). See example 10 (pg 3) of:

    http://www.math.wisc.edu/~robbin/521dir/cont.pdf

    From the same link,f(x) is uniformly contiuous on any interval (Theorem 3, pg 3) if it satisfies a Lipschitz inequality:

    │f(x)-f(x0)│≤ M│x-x0 │< Mδ = ε
    Follow Math Help Forum on Facebook and Google+

  15. #30
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Uniform continuity

    Quote Originally Posted by Hartlw View Post
    The function f(x) = x2 is contiuous but not uniformly continuous on (0,infinity). See example 10 (pg 3) of:
    Of course that is true. But $\displaystyle{\lim _{x \to \infty }}{x^2}$ is not finite.
    Last edited by Plato; February 21st 2014 at 07:02 AM.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 4 FirstFirst 1234 LastLast

Similar Math Help Forum Discussions

  1. Uniform Continuity
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 9th 2011, 04:28 AM
  2. uniform differentiable => uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 30th 2009, 04:19 PM
  3. uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 14th 2009, 06:51 AM
  4. uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: August 17th 2009, 09:37 PM
  5. uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 11th 2009, 08:48 PM

Search Tags


/mathhelpforum @mathhelpforum